Functions help (test soon)

Hey guys, i have a test soon and just wanted to get a few things cleared up. Thanks

1. a. How do u identify the x-intercepts of the graph of y=x(x^2-9a^2) (note: a>0)

b. How would u go about sketching the graph of y= x(x^2-9a^2)

2. If the equation 2x^2 + 6=ax has exactly one solution, find the value(s) of a, such that this is true.

3. Is it possible to find the x intercepts of the function f(x)= 4x^3 - 3x^2- 8x + 5 without using a calculator?

4. If the graph of f: R arrow R crosses the x axis exactly three times, which one of the following rules could not be the rule for f?
A. y=x(x^2-4)
B. y=x(x-2)(x+4)(x^2+1)
C. y=(3-x)(x^4-16)
D y=(x^2-x-6)(x-4)
E y=(x^2-x-12)

5. For the curve of the function with equation y= (2-x)(x-2)(x-5), the subset of R for which the gradient of the graph is positive is best described by?

1. a. How do u identify the x-intercepts of the graph of y=x(x^2-9a^2) (note: a>0)

y = (x+0) (x -3a)(x+3a)

x intercepts are

x+0 = 0 , x = 0
x-3a=0 , x= 3a
x+3a =0 , x= -3a

Originally Posted by chaneliman

b. How would u go about sketching the graph of y=
x(x^2-9a^2)

power of 3 we start with a positive gradient … unless there was a reflection in the x

from left to right pass though -3a 1st then loop back 0 then loop back around to 3a …

Originally Posted by chaneliman

2. If the equation 2x^2 + 6=ax has exactly one solution, find the value(s) of a, such that this is true.

2x^2 –ax + 6 = 0

then take the Discriminate and set it to 0

b^2 - 4ac = 0

a^2 - 4(12)= 0

a^2 = 48

a= +root48 or –root48

Originally Posted by chaneliman

3. Is it possible to find the x intercepts of the function f(x)= 4x^3 - 3x^2- 8x + 5 without using a calculator?

yes … but you will be here all day … and need to know polynomial division which would take a while to teach.. best thing to do .. solve for y=0

Originally Posted by chaneliman

4. If the graph of f: R arrow R crosses the x axis exactly three times, which one of the following rules could not be the rule for f?
A. y=x(x^2-4)
B. y=x(x-2)(x+4)(x^2+1)
C. y=(3-x)(x^4-16)
D y=(x^2-x-6)(x-4)
E y=(x^2-x-12)

E because you can only see power of 2… if you expand it so max amount of crosses is 2 …

Originally Posted by chaneliman

5. For the curve of the function with equation y= (2-x)(x-2)(x-5), the subset of R for which the gradient of the graph is positive is best described by?

ok rearrange y= (-x+2)(x-2)(x-5)

y= -(x-2)(x-2)(x-5)
y= -(x-2)^2(x-5)

Intercepts are … 2 and 5 and 2 is a turning point

ok because its power of the graph is 3 we start positive but we have reflection in x so we actually start negative… we get to 2 bounce up doesn’t not cross remember then we start to get a positive gradient sadly… it returns to pass though 5 … so we need to differentiate the graph…

Assuming that the function is a cubic and assuming (2, 2) is a stationary point of inflexion (earboth, I'm sure that's the word you meant rather than turning point), the following model can be immediately used:

.

To get a, the assumption that (0, 0) is point on the curve will do the trick:

.

Then .

This answer is equivalent to the one found by earboth.

By the way, the point (4, 4) also appears to lie on the curve - our answers are certainly consistent with that.

Mr F says: Well then, bad luck sport. You see, the thing is that has turning points (at x = -1 and x = -1/9). But has a stationary point of inflexion. It's therefore impossible to get from using any combination of translations, dilations or reflection in coordinate axes. So that's that.